package work;

import org.w3c.dom.Node;

import java.util.LinkedList;
import java.util.Queue;

//模拟实现二叉树的功能
public class BinaryTree {

    static class TreeNode {
        public char val;
        public TreeNode left;//左孩子的引用
        public TreeNode right;//右孩子的引用

        public TreeNode(char val) {
            this.val = val;
        }
    }


    /**
     * 创建一棵二叉树 返回这棵树的根节点
     *
     * @return
     */
    public TreeNode createTree() {
        TreeNode a = new TreeNode('A');
        TreeNode b = new TreeNode('B');
        TreeNode c = new TreeNode('C');
        TreeNode d = new TreeNode('D');
        TreeNode e = new TreeNode('E');
        a.left = b;
        a.right = c;
        b.left = d;
        b.right = e;
        return a;
    }

    // 前序遍历
    public void preOrder(TreeNode root) {
        System.out.printf(root.val + "\t");
        postOrder(root.left);
        postOrder(root.right);
    }

    // 中序遍历
    void inOrder(TreeNode root) {
        postOrder(root.left);
        System.out.printf(root.val + "\t");
        postOrder(root.right);
    }

    // 后序遍历
    void postOrder(TreeNode root) {
        postOrder(root.left);
        postOrder(root.right);
        System.out.printf(root.val + "\t");
    }

    public static int nodeSize;

    /**
     * 获取树中节点的个数：遍历思路
     */
    void size(TreeNode root) {


    }

    /**
     * 获取节点的个数：子问题的思路
     *
     * @param root
     * @return
     */
    int size2(TreeNode root) {
        if(root == null){
            return 0;
        }
        int rootSize = 1;
        int leftSize = size2(root.left);
        int rightSize = size2(root.right);
        return rightSize + leftSize +rightSize;
    }


    /*
     获取叶子节点的个数：遍历思路
     */
    public static int leafSize = 0;

    void getLeafNodeCount1(TreeNode root) {
    }

    /*
     获取叶子节点的个数：子问题
     */
    int getLeafNodeCount2(TreeNode root) {
        if(root == null){
            return 0;
        }
        if(root.right == null&&root.left == null){
            return 1;
        }
        return getLeafNodeCount2(root.left) + getLeafNodeCount2(root.right);
    }

    /*
    获取第K层节点的个数
     */
    int getKLevelNodeCount(TreeNode root, int k) {
        if(root == null || k < 0){
            return 0;
        }
        if(k == 1){
            return 1;
        }
        return getKLevelNodeCount(root.left,k - 1) + getKLevelNodeCount(root.right,k - 1);
    }

    /*
     获取二叉树的高度
     时间复杂度：O(N)
     */
    int getHeight(TreeNode root) {
        if(root == null){
            return 0;
        }
        int leftHeight = getHeight(root.left);
        int rightHeight = getHeight(root.right);

        return Math.max(leftHeight,rightHeight) + 1;

    }


    // 检测值为value的元素是否存在
    TreeNode find(TreeNode root, char val) {
        if (root == null) {
            return null;
        }
        if (root.val == val) {
            return root;
        }
        TreeNode leftResult = find(root.left,val);
        if(leftResult != null){
            return leftResult;
        }

        return find(root.right,val);
    }

    //层序遍历
    void levelOrder(TreeNode root) {

    }


    // 判断一棵树是不是完全二叉树
    boolean isCompleteTree(TreeNode root) {
        if(root == null){
            return true;
        }
        boolean isFirstStep = true;
        //判断是第一阶段,还是第二阶段,再对树进行层序遍历
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while(!queue.isEmpty()){
           TreeNode cur = queue.poll();
           if(isFirstStep){
               if(cur.left == null && cur.right == null){
                   isFirstStep = false;
               }else if(cur.left != null && cur.right == null){
                   isFirstStep = false;
               }else if(cur.left == null && cur.right != null){
                   return false;
               }else{
                   queue.offer(cur.left);
                   queue.offer(cur.right);
               }
           }else {
               if (cur.left != null || cur.right != null) {
                   return false;
               }
           }
        }
        return true;
    }
}